Compute the left and right Riemann sums -L4 and R4 respectively - for f(x)=(3-|3-x|) on [0

Compute the left and right Riemann sums -L4 and R4...

Key Concepts

Area Under the Curve

The area under a curve corresponds to the definite integral of the function over a given interval. It quantitatively measures the accumulation of the function’s values over that interval, and Riemann sums provide a method for approximating this area, particularly when an antiderivative is difficult to find or when one is working with discrete data.

Average of Riemann Sums

Averaging the left and right Riemann sums can yield a more accurate approximation of the definite integral by compensating for the over- or under-estimation biases that might occur with either method alone. This average is often closer to the true area under the curve.

Riemann Sums

Riemann sums are a method of approximating the definite integral of a function, which represents the area under its curve. This is achieved by dividing the area into a finite number of rectangles, calculating the area of each rectangle, and then summing these areas to obtain an approximation of the total area.

Left and Right Riemann Sums

Left Riemann sums use the value of the function at the left endpoint of each subinterval to determine the height of the corresponding rectangle, while right Riemann sums use the function’s value at the right endpoint. These two methods can provide different approximations of the integral, especially when the function is increasing or decreasing over the interval.

Partitioning an Interval

Partitioning involves breaking up the overall interval of integration into smaller, equally-spaced subintervals. This allows one to construct rectangles over these subintervals; the width of each subinterval, often denoted as ?x, is a key component in calculating the areas of the rectangles used in the Riemann sums.

Transcript

00:01 Okay, for this problem, we're going to find the left endpoint sum and the right end point sum and the actual area under this curve, f of x, on the interval from 0 to 6.

00:17 So to begin with, we're going to think about how big the interval is, which is 6 minus 0.

00:29 That distance is pretty easy to see is going to be six.

00:32 We're going to divide that up into six sub -intervals, six rectangles.

00:38 And so if we do that, we can see each of our rectangle is going to have a width of one.

00:45 So we'll come over here to the picture and just keep track of that.

00:50 Each of our rectangles will have a width of one.

00:55 And then if we need to think about these x values on the you know, on each of the sub -intervals.

01:04 We're starting at zero.

01:05 That's our first x value.

01:06 And then we could just add one to get our next x value, which would be one, and add another one, which our next x value would be two.

01:15 And our next x value would be three, because we're adding one each time.

01:19 So we're going to use those x values to find our y values.

01:24 And to find our y values, we're going to plug each of those x values into the original problem.

01:30 And so since we're using the left -hand sum, we want to use the very first x value on the left -hand endpoint, and that's a zero.

01:41 So if we plug zero in here, we'll get 3 minus 0 is 3.

01:47 The absolute value of 3 is 3, and 3 minus 3 is 0.

01:50 So our first y value is going to be 0, and i think that's pretty easy to see from the picture.

01:57 So what we're thinking here is the area of that first rectangle has a width of 1, but it has a height of 0, because that's what we just found by evaluating f of 0.

02:17 Now we're going to find our second rectangle, which also has a width of 1, and its height is going to be evaluated at 1.

02:28 So we're going to plug a 1 in here.

02:31 3 minus 1 is 2.

02:33 The absolute value of 2 is 2, and 3 minus 2 is 1.

02:38 We can verify that from the picture.

02:40 It looks like about a height of 1 that we're going to use for this second rectangle right here.

02:48 And now the third rectangle.

02:50 We just need to do 6 of these.

02:53 And they each have a width of 1.

02:55 So i could have used the distributive property with those 1s.

03:00 But we're just going to go ahead and carry on.

03:04 My next x value is 2.